Question 14.5: A graphic equalizer is an audio amplifier that allows you to...

We need to compute values for $R$,$L$,and $C$ that produce a bandpass filter with cutoff frequencies of $1  kHz$   and  $10  kHz$. There  are  many  possible approaches to a solution. For instance, we could use Eqs. 14.29 and 14.30, which specify $\omega _{c1}$ and $\omega _{c2}$ in terms of $R$, $L$, and $C$.

$\omega _{c1} =-\frac{R}{2L} +\sqrt{\left(\frac{R}{2L}^{2}\right) + \left(\frac{1}{LC} \right) }$,

$\omega _{c2} =\frac{R}{2L} +\sqrt{\left(\frac{R}{2L}^{2}\right) + \left(\frac{1}{LC} \right) }$.

Because of the form of these equations, the  algebraic  manipulations  might  get complicated. Instead, we will use the fact that the center frequency is the geometric mean of the cutoff frequencies to compute $\omega _{o}$, and we will then use Eq.14.31 to compute $L$ and $C$ from $\omega _{o}$.

$=\sqrt{\frac{1}{LC} }$.

Next we will use the definition of quality factor to compute $Q$, and last we will use Eq. 14.33 to compute $R$. Even though this approach involves more individual computational steps, each calculation is fairly simple.

$=\sqrt{\frac{L}{CR^{2} } }$.

Any approach we choose will provide only two equations—insufficient  to  solve  for  the  three unknowns—because of the dependencies among the bandpass filter characteristics. Thus, we need to select a value for either $R$,$L$,or $C$ and use the two equations we’ve chosen to calculate the remaining component values. Here, we choose $1  \mu F$ as the capacitor value, because there are stricter limitations on commercially available capacitors than on inductors or resistors.

We compute the center frequency as the geometric mean of the cutoff frequencies:

$f_{0} =\sqrt{f_{c1}f_{c2}} =\sqrt{\left(1000\right)\left(10,000\right) } =3162.28  Hz$.

Next, compute the value of $L$ using the computed center frequency and the selected value for $C$. We must remember to convert the center frequency to radians per second before we can use Eq.14.31:

$L=\frac{1}{\omega ^{2}_{o }C }=\frac{1}{\left[2\pi \left(3162.28\right) \right]^{2} \left(10^{-6} \right) } =2.533  mH$.

The quality factor, $Q$, is defined as the ratio of the center frequency to the bandwidth. The bandwidth is the difference between the two cutoff frequency values. Thus,

$Q=\frac{f_{0}}{f_{c2}-f_{c1}} =\frac{3162.28}{10,000-1000} =0.3514$.

Now use Eq.14.33 to calculate $R$:

$R=\sqrt{\frac{L}{CQ^{2} } } =\sqrt{\frac{0.0025}{\left(10^{-6} \right) \left(0.3514\right) ^{2} } } =143.24  \Omega$.

To check whether these component values produce the bandpass filter we want, substitute them into Eqs.14.29 and 14.30. We find that

$\omega _{c1}=6283.19  rad/ s  \left(1000  Hz\right)$,

$\omega _{c2}=62,831.85  rad/ s  \left(10,000  Hz\right)$,

which are the cutoff frequencies specified for the filter.

This example reminds us that only two of the five bandpass filter parameters can be specified independently. The  other  three  parameters  can always be computed from the two that are specified. In turn, these five parameter values depend on the three component values, $R$, $L$, and $C$, of which only two can be specified independently.